Genetic algorithm calculated values compared to measured
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Saber_phy
el 15 de Jun. de 2018
Comentada: Walter Roberson
el 4 de Oct. de 2018
Hi
I am a student enrolled in the 2nd year physics Master's degree, i am a beginner in Matlab and i have the same problem like Mattias, can you help me and is it was possible to get your e-mail.
My problem is to make a fit of a nonlinear equation to data with the genetic algorithm and to minimize R, my model function is
R*(1-exp(-t/rate))
The industrial data Rexp and t are known: these are two vectors that give the variation of Rexp as a function of time (t)
rate also is known and equal to 40
I declared my function (fitnes function) as follows:
function S=ga_Newton(R)
t=[0 2 4 6 8 10 12 14 16 18 20 22 24 26 28];
Rexp=[1 .7987e-05 3.07996e-05 3.90531e-05 1.89562e-05 2.82794e-05 5.55377e-05
4.75723e-05 6.2447e-05 5.35282e-05 5.16775e-05 4.67144e-05 4.63215e-05 6.62753e-05
3.72551e-05 3.54374e-05];
for i=1:15
S(i) =sum((Rexp(i)-R*(1-exp(-t(i)./40)))^2);
end
The function
S (i) = sum ((Rexp (i) -R * (1-exp (-t (i) ./ 40))) ^ 2)
I use it in the methods of Gauss Newton and Levenberg marquardt and I got a good result of R which is in the order of 0.000059 but with the genetic algorithm I found no result.
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Respuesta aceptada
Walter Roberson
el 15 de Jun. de 2018
Replace
for i=1:15
S(i) =sum((Rexp(i)-R*(1-exp(-t(i)./40)))^2);
end
with
S = sum((Rexp-R*(1-exp(-t./40))).^2);
7 comentarios
Walter Roberson
el 16 de Jun. de 2018
I have attached the exact files I used. Execution requires less than 3 seconds.
More generally, the approach I would use would be to use the Symbolic Toolbox to prepare the function to be minimized:
t=[0 2 4 6 8 10 12 14 16 18 20 22 24 26 28];
Rexp=[1.7987e-05 3.07996e-05 3.90531e-05 1.89562e-05 2.82794e-05 5.55377e-05 4.75723e-05 6.2447e-05 5.35282e-05 5.16775e-05 4.67144e-05 4.63215e-05 6.62753e-05 3.72551e-05 3.54374e-05];
syms R
prediction = R*(1-exp(-t./40));
actual = Rexp;
residue = sum((actual - prediction).^2);
F = matlabFunction(residue);
nvars = 1;
A = []; b = [];
Aeq = []; beq = [];
lb = []; ub = [];
nonlcon = [];
options = gaoptimset('PlotFcn', {@gaplotbestf @gaplotbestindiv @gaplotexpectation @gaplotstopping}, ...
'TolFun', 1e-13, 'Generations', 1000);
[R, fval, exitflag, output] = ga(F, nvars, A, b, Aeq, beq, lb, ub, nonlcon, options);
fprintf('best R: %g\n', R);
fprintf('best residue: %g\n', fval);
Note: you will not get especially close to the true minima. The true minima is at approximately 0.000128381339932429 which can be quickly determined:
[R, fval] = fminunc(F, 0)
Más respuestas (2)
Saber_phy
el 10 de Sept. de 2018
6 comentarios
Walter Roberson
el 13 de Sept. de 2018
"ga does not give an acceptable result within an acceptable time" is a valid conclusion.
You can use ga options to pass in a starting point: look at the initial population matrix parameter.
If it is reasonable to use a lower bound or upper bound on the values, you should do so. Even if it is just lower bound 0.
Saber_phy
el 4 de Oct. de 2018
Editada: Saber_phy
el 4 de Oct. de 2018
3 comentarios
Walter Roberson
el 4 de Oct. de 2018
Use nested functions with shared variables to take a record of all the values you want plotted. Provide your own plot function that you name in the 'PlotFcn' options.
ga does not appear to track individuals as to whether they are converging to the best of the generation. A work around might be for you to use an extra variable in your vector that encoded a unique identity, and code custom mutation and cross-over functions that somehow encoded identity as well, and then provide your own custom plot function that analyzed the Population state to show whatever it is you are trying to plot.
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